\begin{subsection}{Propagación del error para el algoritmo de Machin}
		Para calcular el error de la Fórmula de Machin, primero debemos calcular el error cometido en el algoritmo utilizado para calcular el arcotangente de un número cuyo modulo sea menor que uno.
		
		Sean $t_0, t_1, t_2$ los primeros tres términos sin signo del algoritmo implementado, es decir:
			$$t_k = \dfrac{x^{2k+1}}{2k+1}$$
		Los valores correspondientes a los primeros tres términos de la serie son
			\begin{align*}
				t_0 &= x &
				t_1 &= \dfrac{x^3}{3} &
				t_2 &= \dfrac{x^5}{5}
			\end{align*}

		El cálculo de los primeros tres términos hecho por el algoritmo es \\ $t_0-t_1+t_2$.
		Para analizar el error relativo de los primeros tres términos calculados por el algoritmo analizamos los siguientes errores:
			\newcommand{\errMachTceroTuno}{\dfrac{t_0 \err{t_0} - t_1 \err{t_1}}{t_0-t_1}+\errresta}
%			\newcommand{\errGregTk}[1]{\errdiv^{#1} - \err{2\cdot #1+1}}
			\begin{eqnarray*}
				\err{t_k}     &=& \ferrdiv{x^{2k+1}}{2k+1} \\
				              &=& \ferrpot{x}{(2k+1)} - \err{2k+1} + \errdiv \\
				\err{t_0-t_1} &=& \errMachTceroTuno \label{errMachT0MenosT1} \\
				\err{\arctan(x)}&=& \dfrac{ (t_0-t_1) \err{t_0-t_1} + t_2 \err{t_2} }{ t_0-t_1+t_2 } + \errsuma \label{errMachArgtg} \\
				\err{16\arctan(\frac{1}{5})} &=& \ferrmult{16}{\arctan(\frac{1}{5})} \\
				\err{4\arctan(\frac{1}{239})} &=& \ferrmult{4}{\arctan(\frac{1}{239})} \\
				\err{Machin}  &=& \abs{\ferrresta{16\arctan(\frac{1}{5})}{4\arctan(\frac{1}{239})}}
			\end{eqnarray*}
		El error de cada término del arcotangente se puede acotar por:
			\begin{align*}
				\abs{\err{t_k}} &\leq (2k+1)\abs{\err{x}}+\abs{\ln(x)}(2k+1)\abs{\err{(2k+1)}}+\abs{\errpot} + \abs{\err{2k+1}} + \abs{\errdiv}\\
				&\leq (2k+1)\erep+\abs{\ln(x)}(2k+1)\erep+\erep+\erep+\erep \\
				&\leq \erep( (2k+1)(\abs{\ln(x)}+1) + 3 ) \\
				\abs{\err{t_0}} &\leq \erep( \abs{\ln(x)} + 4 ) \\
				\abs{\err{t_1}} &\leq \erep( 3\abs{\ln(x)} + 6 ) \\
				\abs{\err{t_2}} &\leq \erep( 5\abs{\ln(x)} + 8 ) \\
			\end{align*}
		Luego:
			\begin{align*}
				\abs{\err{t_0-t_1}} &\leq \abs{\errMachTceroTuno}
				\leq \dfrac{t_0 \abs{\err{t_0}} + t_1 \abs{\err{t_1}}}{\abs{t_0-t_1}}+\abs{\errresta} \\
				&\leq \dfrac{t_0\; 2^{1-t} (\abs{\ln(x)}+1) + t_1\; \erep( 3\abs{\ln(x)} + 6 ) }{\abs{t_0-t_1}}+\erep
			\end{align*}

			\begin{align*}
				\abs{\err{\arctan(x)}} &\leq \abs{\dfrac{ (t_0-t_1) \err{t_0-t_1} + t_2 \err{t_2} }{ t_0-t_1+t_2 } + \errsuma } \\
					&\leq \dfrac{ (t_0-t_1) \abs{\err{t_0-t_1}} + t_2 \abs{\err{t_2}} }{t_0-t_1+t_2} + \abs{\errsuma}
					\leq \dfrac{ (t_0-t_1) \abs{\err{t_0-t_1}} + t_2 \abs{\err{t_2}} }{t_0-t_1+t_2} + \erep  \\
					&\leq \dfrac{ \erep (t_0\; (\abs{\ln(x)}+1) + t_1\; ( 3\abs{\ln(x)} + 6 ) + (t_0-t_1))+ t_2 \abs{\err{t_2}} }{t_0-t_1+t_2} + \erep \\
					&\leq \dfrac{ \erep (t_0\; (\abs{\ln(x)}+2) + t_1\; ( 3\abs{\ln(x)} + 5 )+ t_2 ( 5\abs{\ln(x)} + 8 ))}{t_0-t_1+t_2} + \erep \\
					&\leq \dfrac{ \erep (t_0\; (\abs{\ln(x)}+3) + t_1\; ( 3\abs{\ln(x)} + 4 )+ t_2 ( 5\abs{\ln(x)} + 9 ))}{t_0-t_1+t_2} 
			\end{align*}
		Por lo tanto el error de cada arcotangente calulado está acotado por:
			\newcommand{\x}{\frac{1}{5}}
			\begin{align*}	
				\abs{\err{\arctan(\x)}} &\leq \dfrac{ \erep (\x\; (\abs{\ln(\x)}+3) + \frac{(\x)^3}{3} \; ( 3\abs{\ln(\x)} + 4 )+ \frac{(\x)^5}{5} ( 5\abs{\ln(\x)} + 9 ))}{\x-\frac{(\x)^3}{3}+\frac{(\x)^5}{5}} \\
				&\leq \dfrac{ \erep (\x\; (\abs{\ln(\x)}+3) + \frac{(\x)^3}{3} \; ( 3\abs{\ln(\x)} + 4 )+ \frac{(\x)^5}{5} ( 5\abs{\ln(\x)} + 9 ))}{\frac{1}{6}} \\
				&\leq \dfrac{\erep( (\x+(\x)^3+(\x)^5) \abs{\ln(\x)} + \frac{3}{5} + \frac{4}{375} + \frac{9}{15625} )}{\frac{1}{6}} \\
				&\leq \dfrac{\erep( \frac{21}{100} \abs{\ln(\x)} + \frac{31}{50} )}{\frac{1}{6}} 
				\;\leq \dfrac{\erep( \frac{21}{100} \frac{161}{100} + \frac{31}{50} )}{\frac{1}{6}}
				\;\leq \frac{144}{25} \times \erep
			\end{align*}
			
			\renewcommand{\x}{\frac{1}{239}}
			\begin{align*}
				\abs{\err{\arctan{(\x)}}} &\leq \dfrac{\erep( (\x+(\x)^3+(\x)^5) \abs{\ln(\x)} + \frac{4}{239} )}{\frac{2}{239}} \\
				&\leq \dfrac{\erep \frac{4}{239} (\abs{\ln(\x)} + 1 )}{\frac{2}{239}} \\
				&\leq 2^{2-t} (\abs{\ln\left(\x\right)}+1) \leq \frac{31}{20} \times 2^{2-t}
			\end{align*}
		Luego:
			\begin{align*}
				\abs{\err{16\arctan(\frac{1}{5})}} &\leq \abs{\err{16}}+\abs{\err{\arctan(\frac{1}{5})}}+\abs{\errmult} \\
				&\leq 2\times \erep +\frac{144}{25} \times \erep \leq  \frac{194}{25} \times \erep \\
				\abs{\err{4\arctan(\frac{1}{239})}} &\leq \abs{\err{4}}+\abs{\err{\arctan(\frac{1}{239})}}+\abs{\errmult} \\
				&\leq 2\times \erep +\frac{31}{20} \times \erep \leq  \frac{71}{20} \times \erep
			\end{align*}
		Aplicando desigualdad triangular:
			\begin{align*}
				\err{Machin} &\leq \abs{ \frac{{16\arctan(\frac{1}{5})}\err{16\arctan(\frac{1}{5})}-{4\arctan(\frac{1}{239})}\err{4\arctan(\frac{1}{239})}}{16\arctan(\frac{1}{5})-{4\arctan(\frac{1}{239})}} }+ \abs{\errresta} \\
				             &\leq \frac{\abs{16\arctan(\frac{1}{5})\err{16\arctan(\frac{1}{5})} }+\abs{{4\arctan(\frac{1}{239})}\err{4\arctan(\frac{1}{239})}}}{16\arctan(\frac{1}{5})-{4\arctan(\frac{1}{239})}}+ \abs{\errresta} \\
				             &\leq \frac{16\arctan(\frac{1}{5})\abs{\err{16\arctan(\frac{1}{5})} }+4\arctan(\frac{1}{239})\abs{\err{4\arctan(\frac{1}{239})}}}{16\arctan(\frac{1}{5})-{4\arctan(\frac{1}{239})}}+ \abs{\errresta} \\
				             &\leq \frac{ 181 \times \frac{31}{20} \times 2^{2-t} +\frac{24}{25} \times \frac{71}{20} \times \erep }{16\arctan(\frac{1}{5})-{4\arctan(\frac{1}{239})}}+\erep \\ 			 
				             &\leq \frac{ 181 \times \frac{31}{20} \times 2^{2-t} +\frac{24}{25} \times \frac{71}{20} \times \erep }{180}+\erep \\ 			 
				             &\leq \frac{71}{45} \times \erep +\erep \leq \frac{116}{45} \times \erep
			\end{align*}
\end{subsection}
